LeetCode_9. Palindrome Number

9. Palindrome Number

Easy

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121

Output: true

Example 2:

Input: -121

Output: false

Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: 10

Output: false

Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

Follow up:

Coud you solve it without converting the integer to a string?

package leetcode;

public class PalindromeNumber {
	@org.junit.Test
	public void test() {
		int number1 = 121;
		int number2 = -121;
		int number3 = 10;
		PalindromeNumber palindromeNumber = new PalindromeNumber();
		System.out.println(palindromeNumber.isPalindrome(number1));
		System.out.println(palindromeNumber.isPalindrome(number2));
		System.out.println(palindromeNumber.isPalindrome(number3));
	}

	public boolean isPalindrome(int x) {
		// Special cases:
		// As discussed above, when x < 0, x is not a palindrome.
		// Also if the last digit of the number is 0, in order to be a
		// palindrome,
		// the first digit of the number also needs to be 0.
		// Only 0 satisfy this property.
		if (x < 0 || (x % 10 == 0 && x != 0)) {
			return false;
		}

		int revertedNumber = 0;
		while (x > revertedNumber) {
			revertedNumber = revertedNumber * 10 + x % 10;
			x /= 10;
		}

		// When the length is an odd number, we can get rid of the middle digit
		// by revertedNumber/10
		// For example when the input is 12321, at the end of the while loop we
		// get x = 12, revertedNumber = 123,
		// since the middle digit doesn‘t matter in palidrome(it will always
		// equal to itself), we can simply get rid of it.
		return x == revertedNumber || x == revertedNumber / 10;
	}
}

原文地址：https://www.cnblogs.com/denggelin/p/11520815.html